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$PQRS$ is a trapezium in which $PQ \parallel SR$ and its diagonals intersect each other at the point $O$. Show that $\frac{PO}{QO} = \frac{RO}{SO}$.
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In $\Delta POQ$ and $\Delta ROS$
$\angle P = \angle R$
and $\angle Q = \angle S$
$\implies \Delta POQ \sim \Delta ROS$
$\therefore \frac{PO}{RO} = \frac{QO}{SO} \implies \frac{PO}{QO} = \frac{RO}{SO}$
$\angle P = \angle R$
and $\angle Q = \angle S$
$\implies \Delta POQ \sim \Delta ROS$
$\therefore \frac{PO}{RO} = \frac{QO}{SO} \implies \frac{PO}{QO} = \frac{RO}{SO}$